Topic 6 of 8

Eigenvalues and diagonalization

Step 19 · Linear algebra topics

Find the special directions a matrix only stretches, compute eigenvalues from the characteristic equation, and diagonalize a matrix to compute its powers easily.

Eigenvectors and eigenvalues

An eigenvector of A is a nonzero vector that A only stretches: Av = λv. The stretch factor λ is its eigenvalue. Along that direction, A acts like plain multiplication by a number.

A\mathbf{v} = \lambda\mathbf{v}

Example: Checking

Is ⟨1, 1⟩ an eigenvector of $\begin{pmatrix} 3 & 1 \\ 1 & 3 \end{pmatrix}$ ?

A⟨1, 1⟩ = ⟨4, 4⟩.
That is 4⟨1, 1⟩.

AnswerYes, with eigenvalue 4

Example: Another

Is ⟨1, −1⟩ an eigenvector of the same matrix?

A⟨1, −1⟩ = ⟨2, −2⟩, which is 2⟨1, −1⟩.

AnswerYes, with eigenvalue 2

The matrix with rows (3, 1) and (1, 3) stretches ⟨1, 1⟩ by 4 and ⟨1, −1⟩ by 2 (images dashed). Every other direction is turned as well as stretched.

The characteristic equation

Av = λv means (A − λI)v = 0 for some nonzero v, so A − λI must squash space: det(A − λI) = 0. Solve this characteristic equation for λ, then solve (A − λI)v = 0 for each eigenvector.

\det(A - \lambda I) = 0

Example: A 2 × 2 matrix

Find the eigenvalues of $\begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}$ .

det(A − λI) = (4 − λ)(3 − λ) − 2, which is λ² − 7λ + 10.
That factors as (λ − 5)(λ − 2).

Answerλ = 5 and λ = 2

Example: Its eigenvectors

Find an eigenvector for λ = 5.

A − 5I = $\begin{pmatrix} -1 & 1 \\ 2 & -2 \end{pmatrix}$ .
Its first row says −x + y = 0, so y = x.

Answer⟨1, 1⟩

Diagonalization

If A has n independent eigenvectors, put them in the columns of P and the eigenvalues on the diagonal of D. Then A = PDP⁻¹. Powers become easy: Aᵏ = PDᵏP⁻¹, and Dᵏ just raises each diagonal entry to the power k.

A = PDP^{-1} \qquad A^{k} = PD^{k}P^{-1}

Example: Diagonalizing

Diagonalize $\begin{pmatrix} 3 & 1 \\ 1 & 3 \end{pmatrix}$ .

Its eigenvalues are 4 and 2, with eigenvectors ⟨1, 1⟩ and ⟨1, −1⟩.
P = $\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$ and D = $\begin{pmatrix} 4 & 0 \\ 0 & 2 \end{pmatrix}$ .

AnswerA = PDP⁻¹

Example: A power

Use it to find A⁵⟨1, 1⟩.

⟨1, 1⟩ is an eigenvector with eigenvalue 4.
So A⁵ just multiplies it by 4⁵, which is 1024.

Answer⟨1024, 1024⟩

Try it yourself

Work each one on paper first, then open the answer to check.

Is ⟨1, 0⟩ an eigenvector of $\begin{pmatrix} 2 & 5 \\ 0 & 3 \end{pmatrix}$ ?
Show answer
Yes, with eigenvalue 2
Find the eigenvalues of $\begin{pmatrix} 2 & 0 \\ 0 & 7 \end{pmatrix}$ .
Show answer
2 and 7
Find the eigenvalues of $\begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix}$ .
Show answer
3 and −1
A 2 × 2 matrix has trace 6 and determinant 8. What are its eigenvalues?
Show answer
2 and 4
v is an eigenvector of A with eigenvalue 3. What is A²v?
Show answer
9v
Does the 90° rotation matrix have real eigenvalues?
Show answer
No: it turns every direction